am i using a valid alternative solution?

question:

$\displaystyle \sin x+\cos\frac{x}{2}=0$

correct solution:

$\displaystyle \sin(2\cdot\frac{x}{2})+cos\frac{x}{2}=0$

$\displaystyle \2sin\frac{x}{2}\cos\frac{x}{2}+cos\frac{x}{2}=0$

$\displaystyle \cos\frac{x}{2}(\2sin\frac{x}{2}+1)=0$

$\displaystyle \cos\frac{x}{2}=0, \sin\frac{x}{2}=-\frac{1}{2}$

since $\displaystyle 0\leq x<2\pi, \sin\frac{x}{2}\geq 0$

$\displaystyle \sin\frac{x}{2}\neq -\frac{1}{2}$

$\displaystyle x=\pi$

my solution (is this correct?):

$\displaystyle \sin x+\sin(\frac{\pi}{2}\cdot\frac{x}{2})=0$

$\displaystyle 2\sin\frac{x+\pi}{4}=0, \cos\frac{3x-\pi}{4}=0$

$\displaystyle \frac{\pi}{4}\leq\frac{x+\pi}{4}<\frac{3}{4}\pi, /frac{\pi}{4}\leq\frac{3x-\pi}{4}<\frac{5}{4}\pi$

$\displaystyle \frac{x+\pi}{4}=0, \frac{3x-\pi}{4}=\frac{\pi}{2}$

since $\displaystyle 0\leq x<2\pi, x\neq -\pi$

$\displaystyle x=\pi$